Problem 47 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the derivative of the function. $$ y=x 2^{x} $$

Step-by-Step Solution

Answer

The derivative of the function \( y = x2^x \) is \( y' = 2^x + x2^x \ln 2 \)

Step 1: Label the functions

Let u=x and \( v = 2^x \). We will apply the product rule to find the derivative given by \( (uv)' = u'v + uv' \)

Step 2: Find derivative of u and v

The derivative of u is \( u' = 1 \), since the derivative of x with respect to x is 1. The derivative of v is \( v' = 2^x \ln 2 \). The power rule is used here, which states that the derivative of \( a^x \) is \( a^x \ln a \)

Step 3: Apply the product rule

Substitute u, v, u' and v' into the product rule. \( y' = u'v + uv' = 1(2^x) + x(2^x \ln 2) = 2^x + x2^x \ln 2 \)

Key Concepts

Understanding the Product Rule in Calculus

In calculus, finding the derivative of a product of two functions involves the use of the product rule, which is a fundamental concept. Imagine that we have a function made up of two parts, like in our case where \( y = x \cdot 2^x \). Each of these parts can change as \( x \) changes. The product rule helps us find out how quickly the whole function \( y \) changes.

The basic idea is this: if we have two functions multiplied together, say \( u(x) \) and \( v(x) \), the derivative of their product is found using the formula:

• \((uv)' = u'v + uv'\)

This formula combines how fast \( u \) changes (\( u' \)) with how fast \( v \) changes (\( v' \)) to find the overall rate of change.

In our example, we assigned \( u = x \) and \( v = 2^x \). The product rule tells us that to find the derivative of \( y \), we compute \( u'v + uv' \), giving us the complete derivative \( y' = 2^x + x \cdot 2^x \ln(2) \). This formula is great for tackling any scenario where two functions are multiplied, making it an essential tool for calculus students.

Mastering the Power Rule for Derivatives

The power rule is one of the simplest yet powerful tools in calculus used to find derivatives. It specifically applies to functions of the form \( x^n \), where \( n \) is any real number. The rule tells us how to quickly find the derivative of such functions.

The power rule states:

• If \( y = x^n \), then the derivative, \( y' \), is \( nx^{n-1} \).

For instance, if \( y = x^2 \), then according to the power rule, \( y' = 2x^{2-1} = 2x \).

In our exercise, while \( x \) is one function, the other part of the function,\( 2^x \), is actually an exponential function where we apply different rules. This highlights another layer of understanding derivatives: recognizing when to use different rules based on the form of the function we are dealing with. Getting comfortable with this rule builds a solid foundation for tackling more complex derivative problems.

Exploring Exponential Functions

Exponential functions are functions where a constant base is raised to a variable exponent, such as \( a^x \). These functions are unique and important in calculus due to their growth properties. The base \( a \) is a constant and \( x \) is the variable, making these functions scale exponentially as \( x \) changes.

A crucial aspect of exponential functions is their derivative. To find the derivative of \( a^x \), we use the formula:

• \((a^x)' = a^x \ln(a)\)

This formula introduces a natural logarithmic component, which reflects the natural rate of change of these exponential functions.

In our exercise, \( v = 2^x \) is an exponential function where \( a = 2 \). By applying the formula, we found its derivative to be \( v' = 2^x \ln(2) \). The exponential function not only grows faster than polynomial functions, but also its derivative requires understanding logarithmic functions, thus integrating multiple calculus concepts.